NCERT Class 9 Maths Solutions: Polynomials
Question:
Factorise:
x3 + 13x2 + 32x + 20
Concept in a Minute:
Factor Theorem: If p(x) is a polynomial and p(a) = 0, then (x – a) is a factor of p(x).
Remainder Theorem: When a polynomial p(x) is divided by (x – a), the remainder is p(a).
Polynomial Division (Long Division): A method to divide a polynomial by another polynomial of lower degree.
Grouping and Factoring: Techniques used to simplify and factor expressions.
Explanation:
The given polynomial is p(x) = x³ + 13x² + 32x + 20.
We will use the Factor Theorem to find one linear factor. We need to find a value ‘a’ such that p(a) = 0.
We can try integer divisors of the constant term, which is 20. Possible values for ‘a’ are ±1, ±2, ±4, ±5, ±10, ±20.
Let’s test some values:
p(-1) = (-1)³ + 13(-1)² + 32(-1) + 20 = -1 + 13 – 32 + 20 = 0.
Since p(-1) = 0, by the Factor Theorem, (x – (-1)), which is (x + 1), is a factor of p(x).
Now we can perform polynomial division to find the other factor(s). Divide x³ + 13x² + 32x + 20 by (x + 1).
Using long division:
x² + 12x + 20
________________
x + 1 | x³ + 13x² + 32x + 20
-(x³ + x²)
_________
12x² + 32x
-(12x² + 12x)
___________
20x + 20
-(20x + 20)
_________
0
So, x³ + 13x² + 32x + 20 = (x + 1)(x² + 12x + 20).
Now we need to factor the quadratic expression x² + 12x + 20.
We look for two numbers that multiply to 20 and add up to 12. These numbers are 10 and 2.
So, x² + 12x + 20 = x² + 10x + 2x + 20
= x(x + 10) + 2(x + 10)
= (x + 2)(x + 10).
Therefore, the complete factorization of x³ + 13x² + 32x + 20 is (x + 1)(x + 2)(x + 10).
Final Answer: The factors are (x + 1), (x + 2), and (x + 10).
Question:
What are the possible expressions for the dimensions of the cuboids whose volume is given below?
| Volume : 12ky2 + 8ky – 20k |
Concept in a Minute:
The volume of a cuboid is given by the product of its length, breadth, and height. To find the possible expressions for the dimensions of a cuboid given its volume, we need to factorize the expression for the volume into three algebraic expressions.
Explanation:
The given volume of the cuboid is V = 12ky^2 + 8ky – 20k.
To find the possible expressions for the dimensions (length, breadth, and height), we need to factorize this expression.
Step 1: Find the common factor among the terms.
Observe that each term has ‘4k’ as a common factor.
V = 4k(3y^2 + 2y – 5)
Step 2: Factorize the quadratic expression inside the parenthesis.
We need to factorize the quadratic expression 3y^2 + 2y – 5. We can use the splitting the middle term method.
We need to find two numbers whose product is (3 * -5) = -15 and whose sum is +2.
The numbers are +5 and -3.
Step 3: Split the middle term and factor by grouping.
3y^2 + 5y – 3y – 5
Group the terms: (3y^2 + 5y) + (– 3y – 5)
Factor out common factors from each group:
y(3y + 5) – 1(3y + 5)
Now, factor out the common binomial (3y + 5):
(y – 1)(3y + 5)
Step 4: Combine the common factor with the factored quadratic expression.
So, the volume V can be expressed as:
V = 4k * (y – 1) * (3y + 5)
The possible expressions for the dimensions of the cuboid are the factors of the volume. We have three factors: 4k, (y – 1), and (3y + 5).
Therefore, the possible expressions for the dimensions of the cuboid are:
Dimension 1: 4k
Dimension 2: y – 1
Dimension 3: 3y + 5
Note: The order of these dimensions can be interchanged (e.g., length = y-1, breadth = 4k, height = 3y+5, and so on) as the volume is the product of these three dimensions.
Question:
Factorise:
x3 – 2x2 – x + 2
Concept in a Minute:
Polynomial factorization, specifically factoring cubic polynomials. This often involves finding roots (values of x that make the polynomial equal to 0) and using the factor theorem. Grouping terms can also be a useful technique.
Explanation:
The given polynomial is P(x) = x³ – 2x² – x + 2.
Step 1: Try to find a root of the polynomial by testing integer values that are divisors of the constant term (which is 2). The divisors of 2 are ±1, ±2.
Let’s test x = 1:
P(1) = (1)³ – 2(1)² – (1) + 2 = 1 – 2 – 1 + 2 = 0.
Since P(1) = 0, by the Factor Theorem, (x – 1) is a factor of the polynomial.
Step 2: Divide the polynomial by (x – 1) to find the other factors. We can use polynomial long division or synthetic division.
Using synthetic division with root 1:
“`
1 | 1 -2 -1 2
| 1 -1 -2
—————-
1 -1 -2 0
“`
The resulting quadratic is x² – x – 2.
Step 3: Factor the quadratic expression x² – x – 2.
We need two numbers that multiply to -2 and add to -1. These numbers are -2 and 1.
So, x² – x – 2 can be factored as (x – 2)(x + 1).
Step 4: Combine all the factors.
The factors of x³ – 2x² – x + 2 are (x – 1), (x – 2), and (x + 1).
Therefore, the factorised form is (x – 1)(x – 2)(x + 1).
Alternatively, we can try grouping terms:
x³ – 2x² – x + 2
Group the first two terms and the last two terms:
(x³ – 2x²) + (–x + 2)
Factor out common terms from each group:
x²(x – 2) – 1(x – 2)
Now, (x – 2) is a common factor:
(x – 2)(x² – 1)
The term (x² – 1) is a difference of squares, which can be factored as (x – 1)(x + 1).
So, the factorised form is (x – 2)(x – 1)(x + 1).
The final answer is (x – 1)(x – 2)(x + 1).
Question:
Factorise:
x3 – 3x2 – 9x – 5
Concept in a Minute:
Factor Theorem: If P(x) is a polynomial and P(a) = 0, then (x-a) is a factor of P(x).
Polynomial Division: A method to divide a polynomial by another polynomial to find the quotient and remainder.
Explanation:
We are asked to factorise the polynomial P(x) = x³ – 3x² – 9x – 5.
Step 1: Find a root of the polynomial using the Factor Theorem.
We need to find a value ‘a’ such that P(a) = 0. Let’s try integer factors of the constant term (-5), which are ±1, ±5.
Test x = 1:
P(1) = (1)³ – 3(1)² – 9(1) – 5 = 1 – 3 – 9 – 5 = -16 ≠ 0
Test x = -1:
P(-1) = (-1)³ – 3(-1)² – 9(-1) – 5 = -1 – 3(1) + 9 – 5 = -1 – 3 + 9 – 5 = 0
Since P(-1) = 0, by the Factor Theorem, (x – (-1)) which is (x+1) is a factor of P(x).
Step 2: Divide the polynomial by the factor (x+1) using polynomial division or synthetic division.
Using Polynomial Division:
“`
x² -4x -5
________________
x+1 | x³ – 3x² – 9x – 5
-(x³ + x²)
_________
-4x² – 9x
-(-4x² – 4x)
___________
-5x – 5
-(-5x – 5)
_________
0
“`
The quotient is x² – 4x – 5.
Step 3: Factorise the quadratic quotient.
We need to factorise x² – 4x – 5. We look for two numbers that multiply to -5 and add up to -4. These numbers are -5 and 1.
So, x² – 4x – 5 = (x – 5)(x + 1).
Step 4: Combine the factors.
The original polynomial P(x) can be written as the product of its factors:
P(x) = (x + 1) * (x² – 4x – 5)
P(x) = (x + 1) * (x – 5)(x + 1)
P(x) = (x + 1)²(x – 5)
The factorised form of x³ – 3x² – 9x – 5 is (x + 1)²(x – 5).
Question:
Factorise:
2x2 + 7x + 3
Concept in a Minute:
Factorisation of quadratic expressions of the form ax^2 + bx + c. The method involves splitting the middle term (bx) into two terms such that their product equals ac and their sum equals b.
Explanation:
We are asked to factorise the quadratic expression 2x^2 + 7x + 3.
This is a quadratic trinomial of the form ax^2 + bx + c, where a = 2, b = 7, and c = 3.
Step 1: Find the product of ‘a’ and ‘c’.
a * c = 2 * 3 = 6.
Step 2: Find two numbers that multiply to ‘ac’ (which is 6) and add up to ‘b’ (which is 7).
Let these two numbers be p and q. So, p * q = 6 and p + q = 7.
By inspecting factors of 6, we find that 1 * 6 = 6 and 1 + 6 = 7.
So, the two numbers are 1 and 6.
Step 3: Split the middle term (7x) using these two numbers.
We can rewrite 7x as 1x + 6x.
So, the expression becomes: 2x^2 + 1x + 6x + 3.
Step 4: Group the terms and factor out common factors from each group.
Group the first two terms and the last two terms: (2x^2 + x) + (6x + 3).
Factor out the common factor from the first group (2x^2 + x). The common factor is x.
x(2x + 1).
Factor out the common factor from the second group (6x + 3). The common factor is 3.
3(2x + 1).
Step 5: Now the expression is x(2x + 1) + 3(2x + 1).
Notice that (2x + 1) is a common factor in both terms.
Step 6: Factor out the common binomial factor (2x + 1).
(2x + 1)(x + 3).
Therefore, the factorised form of 2x^2 + 7x + 3 is (2x + 1)(x + 3).
To verify, we can expand (2x + 1)(x + 3):
(2x + 1)(x + 3) = 2x(x + 3) + 1(x + 3)
= 2x^2 + 6x + x + 3
= 2x^2 + 7x + 3.
This matches the original expression.
Question:
Factorise:
3x2 – x – 4
Concept in a Minute:
To factorise a quadratic trinomial of the form ax^2 + bx + c, we look for two numbers that multiply to ac and add up to b. Then, we split the middle term (bx) using these two numbers and factor by grouping.
Explanation:
We are asked to factorise the quadratic expression 3x^2 – x – 4.
Here, a = 3, b = -1, and c = -4.
First, we find the product ac: ac = 3 * (-4) = -12.
Next, we need to find two numbers that multiply to -12 and add up to b = -1.
Let’s list the factors of -12:
(1, -12), (-1, 12), (2, -6), (-2, 6), (3, -4), (-3, 4).
Now let’s check the sum of these pairs:
1 + (-12) = -11
-1 + 12 = 11
2 + (-6) = -4
-2 + 6 = 4
3 + (-4) = -1
-3 + 4 = 1
The pair of numbers that satisfy both conditions are 3 and -4, because 3 * (-4) = -12 and 3 + (-4) = -1.
Now, we split the middle term (-x) using these two numbers:
3x^2 – x – 4 = 3x^2 + 3x – 4x – 4.
Next, we factor by grouping:
Group the first two terms and the last two terms:
(3x^2 + 3x) + (– 4x – 4).
Factor out the common factor from each group:
From the first group (3x^2 + 3x), the common factor is 3x:
3x(x + 1).
From the second group (– 4x – 4), the common factor is -4:
-4(x + 1).
Now, the expression becomes:
3x(x + 1) – 4(x + 1).
Notice that (x + 1) is a common factor in both terms. Factor out (x + 1):
(x + 1)(3x – 4).
So, the factorised form of 3x^2 – x – 4 is (x + 1)(3x – 4).
Final Answer: The final answer is $\boxed{(x+1)(3x-4)}$.
Question:
Factorise:
12x2 – 7x + 1
Concept in a Minute:
Factoring quadratic expressions of the form ax^2 + bx + c by splitting the middle term. This involves finding two numbers whose product is ac and whose sum is b.
Explanation:
The given quadratic expression is 12x^2 – 7x + 1.
We need to factorize this expression. We will use the method of splitting the middle term.
Here, a = 12, b = -7, and c = 1.
We need to find two numbers such that their product is a * c = 12 * 1 = 12, and their sum is b = -7.
Let’s list the pairs of factors of 12:
1 and 12 (sum = 13)
2 and 6 (sum = 8)
3 and 4 (sum = 7)
-1 and -12 (sum = -13)
-2 and -6 (sum = -8)
-3 and -4 (sum = -7)
The pair of numbers that satisfies both conditions (product = 12 and sum = -7) is -3 and -4.
Now, we split the middle term (-7x) using these two numbers:
12x^2 – 7x + 1 = 12x^2 – 3x – 4x + 1
Next, we group the terms and factor out common factors from each group:
Group 1: 12x^2 – 3x
The common factor is 3x. So, 12x^2 – 3x = 3x(4x – 1)
Group 2: – 4x + 1
We can factor out -1 from this group to make the expression inside the parenthesis similar to the first group.
– 4x + 1 = -1(4x – 1)
Now, substitute these factored groups back into the expression:
3x(4x – 1) – 1(4x – 1)
Notice that (4x – 1) is a common factor in both terms. We can factor it out:
(4x – 1)(3x – 1)
Thus, the factorization of 12x^2 – 7x + 1 is (4x – 1)(3x – 1).
Final Answer: (4x – 1)(3x – 1)
Question:
Factorise the following using appropriate identity:
9x2 + 6xy + y2
Concept in a Minute:
This question requires the application of algebraic identities. Specifically, the identity for the square of a binomial, (a + b)^2 = a^2 + 2ab + b^2. We need to recognize that the given expression fits this pattern.
Explanation:
The given expression is 9x^2 + 6xy + y^2.
We need to factorise this expression using an appropriate identity.
Let’s look at the identity (a + b)^2 = a^2 + 2ab + b^2.
Compare the given expression with the right-hand side of the identity: a^2 + 2ab + b^2.
The first term of the given expression is 9x^2. We can write this as (3x)^2.
So, we can consider a = 3x.
The last term of the given expression is y^2.
So, we can consider b = y.
Now, let’s check the middle term of the given expression, which is 6xy, and compare it with the middle term of the identity, which is 2ab.
Substitute a = 3x and b = y into 2ab:
2ab = 2 * (3x) * (y) = 6xy.
Since the middle term matches, the given expression 9x^2 + 6xy + y^2 is indeed in the form a^2 + 2ab + b^2, where a = 3x and b = y.
Therefore, we can use the identity (a + b)^2 = a^2 + 2ab + b^2 to factorise the expression.
Substituting a = 3x and b = y into (a + b)^2, we get:
(3x + y)^2.
Thus, the factorised form of 9x^2 + 6xy + y^2 is (3x + y)^2.
This can also be written as (3x + y)(3x + y).
Final Answer: The factorised form of 9x^2 + 6xy + y^2 is (3x + y)^2.
Question:
Factorise the following using appropriate identity:
4y2 – 4y + 1
Concept in a Minute:
The question requires factorizing a quadratic expression. The key concept is recognizing and applying algebraic identities, specifically the identity for a perfect square trinomial: (a – b)2 = a2 – 2ab + b2.
Explanation:
The given expression is 4y2 – 4y + 1.
We need to factorize this expression using an appropriate identity.
Let’s examine the expression to see if it fits any known algebraic identities.
We can observe that the first term, 4y2, is a perfect square, as it can be written as (2y)2.
Similarly, the last term, 1, is also a perfect square, as it can be written as (1)2.
The middle term is -4y. Let’s check if this term can be represented as -2ab, where ‘a’ is the square root of the first term and ‘b’ is the square root of the last term.
Here, a = 2y and b = 1.
So, -2ab = -2 * (2y) * (1) = -4y.
This matches the middle term of the given expression.
Therefore, the expression 4y2 – 4y + 1 is a perfect square trinomial that fits the identity a2 – 2ab + b2 = (a – b)2.
Substituting a = 2y and b = 1 into the identity, we get:
4y2 – 4y + 1 = (2y)2 – 2 * (2y) * (1) + (1)2
= (2y – 1)2
Thus, the factorization of 4y2 – 4y + 1 is (2y – 1)2.
Question:
Factorise the following:
27 – 125a3 – 135a + 225a2
Concept in a Minute:
This question involves factorizing a cubic expression. The key concept is recognizing the pattern of a cubic expansion, specifically (a – b)³ = a³ – 3a²b + 3ab² – b³.
Explanation:
The given expression is 27 – 125a³ – 135a + 225a².
We need to rearrange the terms to see if it fits the pattern of a cubic expansion.
Let’s rearrange it as: -125a³ + 225a² – 135a + 27.
Now, let’s try to identify ‘a’ and ‘b’ in the expansion (a – b)³ = a³ – 3a²b + 3ab² – b³.
Consider the terms with the highest power and the constant term.
We can see that 27 is 3³. So, let’s assume one of the terms is 3.
We can see that 125a³ is (5a)³. So, let’s assume the other term is 5a.
Let’s test if (3 – 5a)³ matches the given expression.
(3 – 5a)³ = 3³ – 3(3²)(5a) + 3(3)(5a)² – (5a)³
= 27 – 3(9)(5a) + 9(25a²) – 125a³
= 27 – 135a + 225a² – 125a³
Rearranging this to match the order of the given expression:
-125a³ + 225a² – 135a + 27
This matches the given expression exactly.
Therefore, the factorization of 27 – 125a³ – 135a + 225a² is (3 – 5a)³.
Alternatively, we could consider (5a – 3)³.
(5a – 3)³ = (5a)³ – 3(5a)²(3) + 3(5a)(3)² – 3³
= 125a³ – 3(25a²)(3) + 15a(9) – 27
= 125a³ – 225a² + 135a – 27
This is the negative of the given expression. So, the factorization is -(5a – 3)³ which is equal to (3 – 5a)³.
The final answer is $\boxed{(3-5a)^3}$.
Question:
Factorise:
6x2 + 5x – 6
Concept in a Minute:
The key concept needed is factorization of a quadratic expression of the form ax^2 + bx + c. This is typically done by splitting the middle term (bx) into two terms such that their product equals ac and their sum equals b.
Explanation:
We need to factorize the quadratic expression 6x^2 + 5x – 6.
First, identify the coefficients: a = 6, b = 5, and c = -6.
Next, calculate the product ac: ac = 6 * (-6) = -36.
Now, we need to find two numbers that multiply to -36 and add up to 5.
Let’s list the factors of -36 and their sums:
-1 * 36 = -36, -1 + 36 = 35
1 * -36 = -36, 1 + (-36) = -35
-2 * 18 = -36, -2 + 18 = 16
2 * -18 = -36, 2 + (-18) = -16
-3 * 12 = -36, -3 + 12 = 9
3 * -12 = -36, 3 + (-12) = -9
-4 * 9 = -36, -4 + 9 = 5
4 * -9 = -36, 4 + (-9) = -5
The two numbers that satisfy the conditions are -4 and 9.
Now, split the middle term (5x) using these two numbers:
6x^2 + 9x – 4x – 6
Next, group the terms and factor out the common factor from each group:
Group 1: 6x^2 + 9x. The common factor is 3x. So, 3x(2x + 3).
Group 2: -4x – 6. The common factor is -2. So, -2(2x + 3).
Now, rewrite the expression with the factored groups:
3x(2x + 3) – 2(2x + 3)
Notice that (2x + 3) is a common factor in both terms. Factor out (2x + 3):
(2x + 3)(3x – 2)
Thus, the factorization of 6x^2 + 5x – 6 is (2x + 3)(3x – 2).
Question:
Following expression is polynomial in one variable or not? State reason for your answer.
4x2 – 3x + 7
Concept in a Minute:
A polynomial in one variable is an algebraic expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. The exponents of the variable(s) must be whole numbers (0, 1, 2, 3, …).
Explanation:
The given expression is 4x^2 – 3x + 7.
Let’s examine the terms in the expression:
1. The first term is 4x^2. Here, the variable is ‘x’ and its exponent is 2. Since 2 is a non-negative integer (a whole number), this term is valid for a polynomial.
2. The second term is -3x. This can be written as -3x^1. Here, the variable is ‘x’ and its exponent is 1. Since 1 is a non-negative integer, this term is valid for a polynomial.
3. The third term is 7. This can be considered as 7x^0, where the variable ‘x’ has an exponent of 0. Since 0 is a non-negative integer, this term is also valid for a polynomial.
All the terms in the expression 4x^2 – 3x + 7 involve the variable ‘x’ raised to non-negative integer powers (2, 1, and 0). The expression only uses addition, subtraction, and multiplication of the variable and coefficients.
Therefore, the expression 4x^2 – 3x + 7 is a polynomial in one variable (x).
Reason: All the exponents of the variable ‘x’ in the expression are non-negative integers.
Question:
Verify:
x3 – y3 = (x – y) (x2 + xy + y2)
Concept in a Minute:
This question involves verifying an algebraic identity. The key concept needed is the distributive property of multiplication over addition, and the ability to expand and simplify polynomial expressions. Specifically, we will be multiplying two binomials (x-y) and a trinomial (x^2 + xy + y^2) to see if we arrive at the expression x^3 – y^3.
Explanation:
To verify the identity x^3 – y^3 = (x – y) (x^2 + xy + y^2), we will expand the right-hand side of the equation using the distributive property.
Step 1: Distribute the ‘x’ term from the first parenthesis (x – y) to each term in the second parenthesis (x^2 + xy + y^2).
x * (x^2 + xy + y^2) = x * x^2 + x * xy + x * y^2
= x^3 + x^2y + xy^2
Step 2: Distribute the ‘-y’ term from the first parenthesis (x – y) to each term in the second parenthesis (x^2 + xy + y^2).
-y * (x^2 + xy + y^2) = -y * x^2 + -y * xy + -y * y^2
= -yx^2 – xy^2 – y^3
= -x^2y – xy^2 – y^3
Step 3: Combine the results from Step 1 and Step 2.
(x^3 + x^2y + xy^2) + (-x^2y – xy^2 – y^3)
Step 4: Group like terms and simplify.
x^3 + (x^2y – x^2y) + (xy^2 – xy^2) – y^3
x^3 + 0 + 0 – y^3
= x^3 – y^3
Since the expanded form of the right-hand side (x – y) (x^2 + xy + y^2) simplifies to x^3 – y^3, which is the left-hand side of the equation, the identity is verified.
Question:
Factorise:
2y3 + y2 – 2y – 1
Concept in a Minute:
Factorization of polynomials, specifically using grouping and identifying common factors. For a cubic polynomial, we might try grouping terms in pairs and factoring out common factors from each pair. If the remaining binomials are the same, we can factor them out. Alternatively, we can use the Factor Theorem to find roots and then factor.
Explanation:
The given polynomial is $2y^3 + y^2 – 2y – 1$.
We can try to factor this polynomial by grouping terms. Let’s group the first two terms and the last two terms:
$(2y^3 + y^2) + (-2y – 1)$
Now, we factor out the greatest common factor from each group.
From the first group, $2y^3 + y^2$, the common factor is $y^2$. Factoring it out, we get:
$y^2(2y + 1)$
From the second group, $-2y – 1$, we can factor out $-1$ to make the expression inside the parenthesis similar to the first group:
$-1(2y + 1)$
Now, we have the expression as:
$y^2(2y + 1) – 1(2y + 1)$
Notice that $(2y + 1)$ is a common factor in both terms. We can factor out $(2y + 1)$:
$(2y + 1)(y^2 – 1)$
The term $(y^2 – 1)$ is a difference of squares, which can be factored as $(y – 1)(y + 1)$.
So, the completely factored form of the polynomial is:
$(2y + 1)(y – 1)(y + 1)$
Therefore, the factorization of $2y^3 + y^2 – 2y – 1$ is $(2y + 1)(y – 1)(y + 1)$.
Question:
Evaluate the following using suitable identity:
(998)3
Concept in a Minute:
The question requires evaluating the cube of a number, specifically 998. Instead of direct multiplication, which would be tedious, we can use algebraic identities to simplify the calculation. The relevant identity here is the cube of a binomial difference: $(a-b)^3 = a^3 – 3a^2b + 3ab^2 – b^3$. We need to express 998 in a form suitable for this identity, such as $(1000-2)$.
Explanation:
We are asked to evaluate $(998)^3$.
We can rewrite 998 as $(1000 – 2)$.
Now, we can use the algebraic identity for the cube of a binomial difference: $(a-b)^3 = a^3 – 3a^2b + 3ab^2 – b^3$.
Here, let $a = 1000$ and $b = 2$.
Substituting these values into the identity:
$(1000 – 2)^3 = (1000)^3 – 3(1000)^2(2) + 3(1000)(2)^2 – (2)^3$
Now, we calculate each term:
$(1000)^3 = 1000 \times 1000 \times 1000 = 1,000,000,000$
$3(1000)^2(2) = 3(1,000,000)(2) = 6,000,000$
$3(1000)(2)^2 = 3(1000)(4) = 12,000$
$(2)^3 = 2 \times 2 \times 2 = 8$
Now, substitute these values back into the expanded form:
$(1000 – 2)^3 = 1,000,000,000 – 6,000,000 + 12,000 – 8$
Perform the subtraction and addition:
$1,000,000,000 – 6,000,000 = 994,000,000$
$994,000,000 + 12,000 = 994,012,000$
$994,012,000 – 8 = 994,011,992$
Therefore, $(998)^3 = 994,011,992$.
Question:
Factorise the following:
64a3 – 27b3 – 144a2b + 108ab2
Concept in a Minute:
The question involves factorizing a cubic expression. The key concept here is recognizing and applying the algebraic identity for the difference of cubes: (x – y)³ = x³ – y³ – 3x²y + 3xy². This identity can be rearranged to group terms in the given expression.
Explanation:
Let’s analyze the given expression: 64a³ – 27b³ – 144a²b + 108ab².
We can try to match this with the expansion of (x – y)³.
Let’s consider x³ = 64a³ and y³ = 27b³.
Taking the cube root of these terms, we get:
x = ³√(64a³) = 4a
y = ³√(27b³) = 3b
Now let’s check the middle terms of the expression with the formula -3x²y + 3xy².
First middle term: -3x²y
Substitute x = 4a and y = 3b:
-3 * (4a)² * (3b) = -3 * (16a²) * (3b) = -3 * 48a²b = -144a²b.
This matches the second term in the given expression.
Second middle term: +3xy²
Substitute x = 4a and y = 3b:
+3 * (4a) * (3b)² = +3 * (4a) * (9b²) = +3 * 36ab² = +108ab².
This matches the third term in the given expression.
Since all terms match the expansion of (x – y)³, we can conclude that the given expression is equal to (4a – 3b)³.
Therefore, the factorization of 64a³ – 27b³ – 144a²b + 108ab² is (4a – 3b)³.
Question:
Verify:
x3 + y3 = (x + y) (x2 – xy + y2)
Concept in a Minute:
This question requires verifying an algebraic identity, specifically the sum of cubes factorization. The key concept is to expand the right-hand side of the equation and show that it simplifies to the left-hand side. This involves the distributive property of multiplication over addition.
Explanation:
To verify the identity x³ + y³ = (x + y) (x² – xy + y²), we will expand the right-hand side of the equation.
Let’s multiply (x + y) by (x² – xy + y²):
(x + y) (x² – xy + y²)
We distribute each term in the first parenthesis to each term in the second parenthesis:
x * (x² – xy + y²) + y * (x² – xy + y²)
Now, perform the multiplication for each part:
(x * x² – x * xy + x * y²) + (y * x² – y * xy + y * y²)
Simplify each term:
(x³ – x²y + xy²) + (x²y – xy² + y³)
Now, combine like terms. We can see that -x²y and +x²y cancel each other out. Similarly, +xy² and -xy² cancel each other out.
x³ + (-x²y + x²y) + (xy² – xy²) + y³
This simplifies to:
x³ + 0 + 0 + y³
Which is:
x³ + y³
Since the expanded form of the right-hand side (x + y) (x² – xy + y²) equals the left-hand side (x³ + y³), the identity is verified.
Question:
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Concept in a Minute:
Algebraic identity: x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
Factorization of (a+b+c)³ (less common, but can be used as an alternative approach)
Explanation:
Given the condition x + y + z = 0.
We need to show that x³ + y³ + z³ = 3xyz.
Let’s use the algebraic identity:
x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
Substitute the given condition x + y + z = 0 into the identity:
x³ + y³ + z³ – 3xyz = (0)(x² + y² + z² – xy – yz – zx)
Any expression multiplied by zero is zero. Therefore:
x³ + y³ + z³ – 3xyz = 0
Now, rearrange the equation to isolate x³ + y³ + z³:
x³ + y³ + z³ = 3xyz
This proves the given statement.
Alternative Approach using substitution:
From x + y + z = 0, we can write z = -(x + y).
Now substitute this into the expression x³ + y³ + z³:
x³ + y³ + (-(x + y))³
= x³ + y³ – (x + y)³
We know the identity for (x + y)³: (x + y)³ = x³ + y³ + 3xy(x + y).
So, x³ + y³ – (x³ + y³ + 3xy(x + y))
= x³ + y³ – x³ – y³ – 3xy(x + y)
= -3xy(x + y)
Now, recall that z = -(x + y). So, substitute -(x + y) back with z:
= -3xy(-z)
= 3xyz
Thus, x³ + y³ + z³ = 3xyz.
Question:
Factorise:
27x3 + y3 + z3 – 9xyz
Concept in a Minute:
The question requires factorization of a cubic expression. The key concept involved is the algebraic identity: a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca).
Explanation:
The given expression is 27x³ + y³ + z³ – 9xyz.
We can rewrite 27x³ as (3x)³.
So, the expression becomes (3x)³ + y³ + z³ – 9xyz.
This expression matches the form a³ + b³ + c³ – 3abc, where:
a = 3x
b = y
c = z
Now, we need to check if the term –9xyz corresponds to –3abc.
–3abc = –3 * (3x) * y * z = –9xyz.
This matches the given expression.
Therefore, we can apply the identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca).
Substituting the values of a, b, and c:
a + b + c = 3x + y + z
a² = (3x)² = 9x²
b² = y²
c² = z²
ab = (3x) * y = 3xy
bc = y * z = yz
ca = z * (3x) = 3zx
So, a² + b² + c² – ab – bc – ca = 9x² + y² + z² – 3xy – yz – 3zx.
Therefore, the factorization of 27x³ + y³ + z³ – 9xyz is:
(3x + y + z)(9x² + y² + z² – 3xy – yz – 3zx).
Final Answer: The factorization of 27x³ + y³ + z³ – 9xyz is (3x + y + z)(9x² + y² + z² – 3xy – yz – 3zx).
Question:
Find the zero of the polynomial in the following case:
p(x) = ax, a ≠ 0
Concept in a Minute:
A zero of a polynomial p(x) is a value of x for which p(x) = 0. To find the zero of a polynomial, we set the polynomial equal to zero and solve for x.
Explanation:
We are given the polynomial p(x) = ax, where it is specified that a ≠ 0.
To find the zero of this polynomial, we need to find the value of x for which p(x) = 0.
So, we set the polynomial equal to zero:
ax = 0
Now, we need to solve this equation for x.
Since we are given that a ≠ 0, we can divide both sides of the equation by ‘a’ without any issue of dividing by zero.
Dividing both sides by ‘a’:
(ax) / a = 0 / a
x = 0
Therefore, the zero of the polynomial p(x) = ax (where a ≠ 0) is x = 0.
Step-by-step solution:
1. Understand the definition of a zero of a polynomial: A value of x that makes the polynomial equal to zero.
2. Set the given polynomial p(x) equal to zero: ax = 0.
3. Analyze the given condition: a ≠ 0.
4. Solve the equation ax = 0 for x. Since a is not zero, divide both sides by a.
5. x = 0 / a
6. x = 0.
7. The zero of the polynomial p(x) = ax is 0.
Question:
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case:
p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
Concept in a Minute:
The Factor Theorem states that a polynomial p(x) has a factor (x – a) if and only if p(a) = 0. In simpler terms, if substituting a value ‘a’ into the polynomial results in zero, then (x – a) is a factor of that polynomial. Conversely, if (x – a) is a factor of p(x), then p(a) must be equal to 0.
Explanation:
We are given two polynomials: p(x) = x³ + 3x² + 3x + 1 and g(x) = x + 2.
We need to determine if g(x) is a factor of p(x) using the Factor Theorem.
According to the Factor Theorem, if g(x) is a factor of p(x), then the root of g(x) must also be a root of p(x).
First, find the root of g(x). To do this, set g(x) equal to zero and solve for x:
g(x) = x + 2 = 0
x = -2
Now, substitute this value of x (-2) into p(x) and evaluate p(-2):
p(x) = x³ + 3x² + 3x + 1
p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1
p(-2) = -8 + 3(4) – 6 + 1
p(-2) = -8 + 12 – 6 + 1
p(-2) = 4 – 6 + 1
p(-2) = -2 + 1
p(-2) = -1
Since p(-2) is not equal to 0 (p(-2) = -1), according to the Factor Theorem, g(x) = x + 2 is not a factor of p(x) = x³ + 3x² + 3x + 1.
Question:
Find the value of the polynomial 5x – 4x2 + 3 at x = –1.
Concept in a Minute:
Evaluating a polynomial means substituting a given value for the variable and then simplifying the expression using the order of operations (PEMDAS/BODMAS).
Explanation:
We are asked to find the value of the polynomial 5x – 4x^2 + 3 when x = –1.
Step 1: Write down the given polynomial.
P(x) = 5x – 4x^2 + 3
Step 2: Substitute the given value of x (which is –1) into the polynomial.
P(–1) = 5(–1) – 4(–1)^2 + 3
Step 3: Simplify the expression by performing the operations in the correct order (parentheses, exponents, multiplication/division, addition/subtraction).
First, calculate the exponent:
(–1)^2 = (–1) * (–1) = 1
Now substitute this back into the expression:
P(–1) = 5(–1) – 4(1) + 3
Next, perform the multiplications:
5(–1) = –5
4(1) = 4
Substitute these values back into the expression:
P(–1) = –5 – 4 + 3
Finally, perform the addition and subtraction from left to right:
–5 – 4 = –9
–9 + 3 = –6
So, the value of the polynomial 5x – 4x^2 + 3 at x = –1 is –6.
Final Answer: The final answer is $\boxed{-6}$
Question:
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case:
p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1
Concept in a Minute:
The Factor Theorem states that if p(x) is a polynomial and (x – a) is a linear factor, then p(a) = 0 if and only if (x – a) is a factor of p(x). In simpler terms, if substituting the root of the potential factor into the polynomial results in zero, then the potential factor is indeed a factor of the polynomial.
Explanation:
To determine if g(x) = x + 1 is a factor of p(x) = 2x3 + x2 – 2x – 1, we need to use the Factor Theorem.
Step 1: Find the root of the potential factor g(x).
Set g(x) = 0:
x + 1 = 0
Solving for x, we get:
x = -1
Step 2: Substitute this root into the polynomial p(x).
We need to evaluate p(-1):
p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1
Step 3: Calculate the value of p(-1).
p(-1) = 2(-1) + 1 – (-2) – 1
p(-1) = -2 + 1 + 2 – 1
p(-1) = 0
Step 4: Apply the Factor Theorem.
Since p(-1) = 0, according to the Factor Theorem, (x – (-1)), which is (x + 1), is a factor of p(x).
Therefore, g(x) = x + 1 is a factor of p(x) = 2x3 + x2 – 2x – 1.
Question:
Evaluate the following product without multiplying directly:
104 × 96
Concept in a Minute:
The key concept needed is the algebraic identity: (a + b)(a – b) = a² – b². This identity allows us to simplify the multiplication of two numbers by expressing them as a sum and difference of another number.
Explanation:
The problem asks us to evaluate 104 × 96 without direct multiplication. We can rewrite the numbers 104 and 96 in a way that fits the algebraic identity (a + b)(a – b) = a² – b².
We observe that both 104 and 96 are close to 100.
We can express 104 as (100 + 4).
We can express 96 as (100 – 4).
Now, the product 104 × 96 can be written as (100 + 4) × (100 – 4).
This perfectly matches the form (a + b)(a – b), where a = 100 and b = 4.
Using the identity (a + b)(a – b) = a² – b², we can substitute the values of a and b:
(100 + 4)(100 – 4) = 100² – 4²
Now, we evaluate the squares:
100² = 100 × 100 = 10000
4² = 4 × 4 = 16
Substitute these values back into the equation:
10000 – 16
Finally, perform the subtraction:
10000 – 16 = 9984
Therefore, 104 × 96 = 9984.
Question:
Evaluate the following using suitable identity:
(102)3
Concept in a Minute:
The question requires evaluating (102)^3. This can be done efficiently using the algebraic identity (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3. The number 102 can be broken down into a sum of two convenient numbers, like 100 + 2, to apply this identity.
Explanation:
To evaluate (102)^3 using a suitable identity, we can rewrite 102 as the sum of two numbers whose cube and square are easy to calculate. Let’s choose 102 = 100 + 2.
We will use the algebraic identity: (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3.
In this case, let a = 100 and b = 2.
Substitute these values into the identity:
(100 + 2)^3 = (100)^3 + 3(100)^2(2) + 3(100)(2)^2 + (2)^3
Now, calculate each term:
(100)^3 = 100 * 100 * 100 = 1,000,000
3(100)^2(2) = 3 * (100 * 100) * 2 = 3 * 10,000 * 2 = 60,000
3(100)(2)^2 = 3 * 100 * (2 * 2) = 3 * 100 * 4 = 1,200
(2)^3 = 2 * 2 * 2 = 8
Now, add all the calculated terms:
1,000,000 + 60,000 + 1,200 + 8 = 1,061,208
Therefore, (102)^3 = 1,061,208.
Question:
Factorise the following:
8a3 + b3 + 12a2b + 6ab2
Concept in a Minute:
This problem involves factorizing a cubic expression. The key concept here is recognizing and applying the algebraic identity for the cube of a binomial: (x + y)³ = x³ + y³ + 3x²y + 3xy². We need to see if the given expression fits this pattern.
Explanation:
The given expression is 8a³ + b³ + 12a²b + 6ab².
Let’s try to match this with the expansion of (x + y)³ = x³ + y³ + 3x²y + 3xy².
We can observe that 8a³ can be written as (2a)³.
And b³ is already in the form of a cube.
So, let’s consider x = 2a and y = b.
Now let’s check the middle terms of the expansion:
3x²y = 3 * (2a)² * b = 3 * (4a²) * b = 12a²b. This matches the third term in the given expression.
3xy² = 3 * (2a) * b² = 6ab². This matches the fourth term in the given expression.
Since all the terms match the expansion of (x + y)³ with x = 2a and y = b, we can factorize the given expression as (2a + b)³.
Therefore, the factorization of 8a³ + b³ + 12a²b + 6ab² is (2a + b)³.
Question:
Factorise the following:
64m3 – 343n3
Concept in a Minute:
The question requires factorization of a difference of cubes. The formula for the difference of cubes is $a^3 – b^3 = (a – b)(a^2 + ab + b^2)$.
Explanation:
The given expression is $64m^3 – 343n^3$.
We need to identify the terms that can be expressed as cubes.
We know that $64 = 4^3$ and $343 = 7^3$.
Therefore, $64m^3$ can be written as $(4m)^3$ and $343n^3$ can be written as $(7n)^3$.
So, the expression becomes $(4m)^3 – (7n)^3$.
This is in the form $a^3 – b^3$, where $a = 4m$ and $b = 7n$.
Using the difference of cubes formula, $a^3 – b^3 = (a – b)(a^2 + ab + b^2)$:
Substitute the values of $a$ and $b$:
$(4m)^3 – (7n)^3 = (4m – 7n)((4m)^2 + (4m)(7n) + (7n)^2)$
Now, simplify the terms inside the second parenthesis:
$(4m)^2 = 16m^2$
$(4m)(7n) = 28mn$
$(7n)^2 = 49n^2$
So, the factored expression is:
$(4m – 7n)(16m^2 + 28mn + 49n^2)$
Therefore, the factorization of $64m^3 – 343n^3$ is $(4m – 7n)(16m^2 + 28mn + 49n^2)$.
Question:
Factorise the following:
8a3 – b3 – 12a2b + 6ab2
Concept in a Minute:
The problem requires factorization of a cubic expression. The expression resembles the expansion of a binomial cube, specifically (x – y)^3 = x^3 – y^3 – 3x^2y + 3xy^2. Recognizing this pattern will be key to solving the problem efficiently.
Explanation:
The given expression is 8a^3 – b^3 – 12a^2b + 6ab^2.
We can rewrite this expression to match the form of the binomial cube expansion.
Notice that 8a^3 can be written as (2a)^3 and b^3 is already in the cubic form.
Let’s compare the middle terms with the expansion of (x – y)^3 = x^3 – y^3 – 3x^2y + 3xy^2.
If we set x = 2a and y = b, then:
x^3 = (2a)^3 = 8a^3
y^3 = b^3
-3x^2y = -3(2a)^2(b) = -3(4a^2)(b) = -12a^2b
+3xy^2 = +3(2a)(b)^2 = +6ab^2
The given expression exactly matches the expansion of (2a – b)^3.
Therefore, 8a^3 – b^3 – 12a^2b + 6ab^2 = (2a – b)^3.
To further factorize, we can write (2a – b)^3 as (2a – b)(2a – b)(2a – b).
The final answer is (2a – b)^3.
Question:
Evaluate the following using suitable identity:
(99)3
Concept in a Minute:
This question requires the application of algebraic identities, specifically the cube of a binomial identity. The identity (a – b)³ = a³ – 3a²b + 3ab² – b³ is crucial for simplifying the calculation of the cube of a number that is close to a round number.
Explanation:
We need to evaluate (99)³.
We can rewrite 99 as (100 – 1).
Now, we can use the algebraic identity (a – b)³ = a³ – 3a²b + 3ab² – b³.
Here, a = 100 and b = 1.
Substituting these values into the identity:
(100 – 1)³ = (100)³ – 3(100)²(1) + 3(100)(1)² – (1)³
Calculate each term:
(100)³ = 100 × 100 × 100 = 1,000,000
3(100)²(1) = 3 × (100 × 100) × 1 = 3 × 10,000 = 30,000
3(100)(1)² = 3 × 100 × 1 = 300
(1)³ = 1
Now, substitute these calculated values back into the expanded form:
(99)³ = 1,000,000 – 30,000 + 300 – 1
Perform the arithmetic operations:
1,000,000 – 30,000 = 970,000
970,000 + 300 = 970,300
970,300 – 1 = 970,299
Therefore, (99)³ = 970,299.
Question:
Find the value of the polynomial 5x – 4x2 + 3 at x = 2.
Concept in a Minute:
A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Evaluating a polynomial means substituting a given value for the variable and simplifying the expression.
Explanation:
We are asked to find the value of the polynomial 5x – 4x^2 + 3 when x = 2.
To do this, we substitute the value 2 for x in the polynomial expression.
Step 1: Write down the given polynomial.
Polynomial: P(x) = 5x – 4x^2 + 3
Step 2: Substitute x = 2 into the polynomial.
P(2) = 5(2) – 4(2)^2 + 3
Step 3: Perform the multiplication and exponentiation.
P(2) = 10 – 4(4) + 3
Step 4: Perform the multiplication again.
P(2) = 10 – 16 + 3
Step 5: Perform the addition and subtraction from left to right.
P(2) = -6 + 3
P(2) = -3
Therefore, the value of the polynomial 5x – 4x^2 + 3 at x = 2 is -3.
Question:
Expand the following, using suitable identity:
(–2x + 3y + 2z)2
Concept in a Minute:
The question requires expanding a trinomial squared. The relevant algebraic identity is $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$. This identity allows us to expand the square of a sum of three terms.
Explanation:
We need to expand the expression (–2x + 3y + 2z)². We will use the algebraic identity $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
In this case, we can identify:
a = –2x
b = 3y
c = 2z
Now, we substitute these values into the identity:
(–2x + 3y + 2z)² = (–2x)² + (3y)² + (2z)² + 2(–2x)(3y) + 2(3y)(2z) + 2(2z)(–2x)
Let’s calculate each term:
(–2x)² = (–2)² * x² = 4x²
(3y)² = 3² * y² = 9y²
(2z)² = 2² * z² = 4z²
2(–2x)(3y) = 2 * (–6xy) = –12xy
2(3y)(2z) = 2 * (6yz) = 12yz
2(2z)(–2x) = 2 * (–4zx) = –8zx
Now, we combine all these terms:
(–2x + 3y + 2z)² = 4x² + 9y² + 4z² – 12xy + 12yz – 8zx
Therefore, the expanded form of (–2x + 3y + 2z)² is 4x² + 9y² + 4z² – 12xy + 12yz – 8zx.
Question:
Use suitable identity to find the following product:
(3x + 4) (3x – 5)
Concept in a Minute:
The question requires the use of a suitable algebraic identity to find the product of two binomials. Specifically, the identity (x + a)(x + b) = x² + (a + b)x + ab is relevant here, where x is a common term in both binomials.
Explanation:
The given expression is (3x + 4) (3x – 5).
We can observe that this expression is in the form of (x + a)(x + b), where:
x = 3x
a = 4
b = -5
Now, we can apply the identity (x + a)(x + b) = x² + (a + b)x + ab.
Substitute the values of x, a, and b into the identity:
(3x)² + (4 + (-5)) (3x) + (4)(-5)
First, calculate the square of x:
(3x)² = 3² * x² = 9x²
Next, calculate the sum of a and b and multiply it by x:
(4 + (-5)) (3x) = (-1) (3x) = -3x
Finally, calculate the product of a and b:
(4)(-5) = -20
Now, combine these terms to get the final product:
9x² + (-3x) + (-20)
9x² – 3x – 20
Therefore, the product of (3x + 4) (3x – 5) is 9x² – 3x – 20.
Question:
Use suitable identity to find the following product:
(x + 8) (x – 10)
Concept in a Minute:
The question requires finding the product of two binomials of the form (x + a)(x + b). The suitable identity to use is (x + a)(x + b) = x^2 + (a + b)x + ab.
Explanation:
We are asked to find the product of (x + 8) and (x – 10) using a suitable identity.
The given expression is in the form (x + a)(x + b), where a = 8 and b = –10.
The standard identity for the product of two binomials of this form is:
(x + a)(x + b) = x^2 + (a + b)x + ab
Now, substitute the values of a and b into the identity:
a = 8
b = –10
(x + 8)(x – 10) = x^2 + (8 + (–10))x + (8)(–10)
First, calculate the sum of a and b:
8 + (–10) = 8 – 10 = –2
Next, calculate the product of a and b:
(8)(–10) = –80
Now, substitute these values back into the identity:
(x + 8)(x – 10) = x^2 + (–2)x + (–80)
Simplify the expression:
(x + 8)(x – 10) = x^2 – 2x – 80
Therefore, the product of (x + 8)(x – 10) is x^2 – 2x – 80.
Question:
Determine the following polynomial has (x + 1) a factor:
x4 + 3x3 + 3x2 + x + 1
Concept in a Minute:
Factor Theorem: A polynomial p(x) has (x – a) as a factor if and only if p(a) = 0. In this case, we are checking for the factor (x + 1), which can be written as (x – (-1)). Therefore, we need to check if the polynomial evaluates to 0 when x = -1.
Explanation:
Let the given polynomial be P(x) = x⁴ + 3x³ + 3x² + x + 1.
To determine if (x + 1) is a factor, we use the Factor Theorem.
According to the Factor Theorem, (x + 1) is a factor of P(x) if P(-1) = 0.
Now, let’s substitute x = -1 into the polynomial P(x):
P(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1
Calculate each term:
(-1)⁴ = 1
3(-1)³ = 3(-1) = -3
3(-1)² = 3(1) = 3
(-1) = -1
+ 1 = 1
Now, add all the calculated terms:
P(-1) = 1 – 3 + 3 – 1 + 1
Combine the terms:
P(-1) = (1 + 3 + 1) + (-3 – 1)
P(-1) = 5 – 4
P(-1) = 1
Since P(-1) = 1, which is not equal to 0, (x + 1) is not a factor of the given polynomial.
Question:
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Concept in a Minute:
A monomial is an algebraic expression consisting of a single term. A binomial is an algebraic expression consisting of two terms. The degree of a polynomial is the highest power of the variable in the expression.
Explanation:
To form a binomial of degree 35, we need an expression with two terms, where the highest power of the variable is 35. For example, we can use the variable x. One term could be $x^{35}$ and another term could be a constant, like 5. So, a binomial of degree 35 could be $x^{35} + 5$.
To form a monomial of degree 100, we need an expression with a single term, where the highest power of the variable is 100. Using the variable x, a monomial of degree 100 could be $x^{100}$. Another example could be $7x^{100}$.
Therefore, one example of a binomial of degree 35 is $x^{35} + 5$.
One example of a monomial of degree 100 is $x^{100}$.
Question:
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case:
p(x) = x3 − 4x2 + x + 6, g(x) = x − 3
Concept in a Minute:
The Factor Theorem states that for a polynomial p(x), if p(a) = 0, then (x – a) is a factor of p(x). Conversely, if (x – a) is a factor of p(x), then p(a) = 0. To determine if g(x) is a factor of p(x), we need to find the root of g(x) and then evaluate p(x) at that root. If the result is 0, then g(x) is a factor.
Explanation:
We are given p(x) = x³ − 4x² + x + 6 and g(x) = x − 3.
To use the Factor Theorem, we first need to find the root of g(x). We set g(x) = 0 and solve for x:
x − 3 = 0
x = 3
Now, we need to evaluate p(x) at x = 3. This means substituting 3 for every x in the expression for p(x):
p(3) = (3)³ − 4(3)² + (3) + 6
p(3) = 27 − 4(9) + 3 + 6
p(3) = 27 − 36 + 3 + 6
p(3) = 36 − 36
p(3) = 0
Since p(3) = 0, according to the Factor Theorem, (x – 3), which is g(x), is a factor of p(x).
Therefore, g(x) is a factor of p(x).
Question:
Expand the following, using suitable identity:
(2x – y + z)2
Concept in a Minute:
The question requires expanding a trinomial squared. The suitable identity for this is (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca. We need to carefully identify ‘a’, ‘b’, and ‘c’ from the given expression and substitute them into the identity, paying close attention to the signs.
Explanation:
We are asked to expand (2x – y + z)^2.
We can use the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.
In our expression, let:
a = 2x
b = -y
c = z
Now, substitute these values into the identity:
(2x – y + z)^2 = (2x)^2 + (-y)^2 + (z)^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
Let’s expand each term:
(2x)^2 = 4x^2
(-y)^2 = y^2
(z)^2 = z^2
2(2x)(-y) = -4xy
2(-y)(z) = -2yz
2(z)(2x) = 4zx
Now, combine all the expanded terms:
4x^2 + y^2 + z^2 – 4xy – 2yz + 4zx
Therefore, (2x – y + z)^2 = 4x^2 + y^2 + z^2 – 4xy – 2yz + 4zx.
Question:
Write the coefficient of x2 in the following:-
2 – x2 + x3
Concept in a Minute:
A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. In a polynomial, the coefficient of a term is the number that multiplies the variable part of that term. When identifying the coefficient of x^2, we look for the number directly preceding and multiplying the x^2 term. If there is no explicit number, it is understood to be 1.
Explanation:
The given expression is 2 – x^2 + x^3.
We need to find the coefficient of x^2.
Let’s examine each term in the expression:
The first term is 2. This is a constant term and does not have an x^2 component.
The second term is -x^2. This term has x^2. The number multiplying x^2 is -1.
The third term is x^3. This term has x^3 and not x^2.
Therefore, the coefficient of x^2 in the expression 2 – x^2 + x^3 is -1.
Question:
Use suitable identity to find the following product:
(x + 4) (x + 10)
Concept in a Minute:
The question requires finding the product of two binomials of the form (x + a)(x + b). The suitable identity to use is the distributive property or the specific algebraic identity derived from it: (x + a)(x + b) = x^2 + (a + b)x + ab.
Explanation:
We are asked to find the product of (x + 4) and (x + 10).
We can use the algebraic identity: (x + a)(x + b) = x^2 + (a + b)x + ab.
In this case, a = 4 and b = 10.
Substitute these values into the identity:
(x + 4)(x + 10) = x^2 + (4 + 10)x + (4 * 10)
= x^2 + (14)x + 40
= x^2 + 14x + 40
Alternatively, we can use the distributive property (also known as FOIL):
(x + 4)(x + 10) = x(x + 10) + 4(x + 10)
= x*x + x*10 + 4*x + 4*10
= x^2 + 10x + 4x + 40
Combine the like terms (10x and 4x):
= x^2 + (10 + 4)x + 40
= x^2 + 14x + 40
The final answer is x^2 + 14x + 40.
Question:
Expand the following, using suitable identity:
(x + 2y + 4z)2
Concept in a Minute:
The question requires expanding a trinomial squared. This can be solved using the algebraic identity for the square of a trinomial: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
Explanation:
We need to expand $(x + 2y + 4z)^2$.
We can use the identity $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
In this case, let $a = x$, $b = 2y$, and $c = 4z$.
Substitute these values into the identity:
$(x + 2y + 4z)^2 = (x)^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$
Now, calculate each term:
$(x)^2 = x^2$
$(2y)^2 = 2^2 \times y^2 = 4y^2$
$(4z)^2 = 4^2 \times z^2 = 16z^2$
$2(x)(2y) = 4xy$
$2(2y)(4z) = 16yz$
$2(4z)(x) = 8zx$
Combine all the terms:
$(x + 2y + 4z)^2 = x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$
So, the expanded form of $(x + 2y + 4z)^2$ is $x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$.
Question:
Evaluate the following product without multiplying directly:
95 × 96
Concept in a Minute:
The question requires evaluating a product without direct multiplication. This suggests using algebraic identities to simplify the calculation. Specifically, the identity (a – b)(c – d) or similar forms involving adjustments to standard numbers will be useful. We can rewrite the numbers in terms of a common base close to them, like 100.
Explanation:
We are asked to evaluate 95 × 96 without direct multiplication.
We can rewrite the numbers as:
95 = 100 – 5
96 = 100 – 4
Now, the product becomes:
(100 – 5) × (100 – 4)
We can use the algebraic identity: (x – a)(x – b) = x² – (a + b)x + ab
In this case, x = 100, a = 5, and b = 4.
Applying the identity:
(100 – 5)(100 – 4) = (100)² – (5 + 4) × 100 + (5 × 4)
= 10000 – (9) × 100 + 20
= 10000 – 900 + 20
= 9100 + 20
= 9120
Therefore, 95 × 96 = 9120.
Question:
Verify whether the following zeroes of the polynomial, indicated against them.
p(x) = x2 – 1, x = 1, –1
Concept in a Minute:
A zero of a polynomial p(x) is a value of x for which p(x) = 0. To verify if a given value is a zero, substitute that value into the polynomial and check if the result is zero.
Explanation:
We are given the polynomial p(x) = x2 – 1 and we need to verify if x = 1 and x = –1 are its zeroes.
To verify if x = 1 is a zero, we substitute x = 1 into the polynomial p(x):
p(1) = (1)2 – 1
p(1) = 1 – 1
p(1) = 0
Since p(1) = 0, x = 1 is a zero of the polynomial p(x) = x2 – 1.
To verify if x = –1 is a zero, we substitute x = –1 into the polynomial p(x):
p(–1) = (–1)2 – 1
p(–1) = 1 – 1
p(–1) = 0
Since p(–1) = 0, x = –1 is a zero of the polynomial p(x) = x2 – 1.
Therefore, both x = 1 and x = –1 are zeroes of the polynomial p(x) = x2 – 1.
Question:
Expand the following, using suitable identity:
(–2x + 5y – 3z)2
Concept in a Minute:
The question requires expanding a trinomial squared. The suitable identity for this is the square of a trinomial: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. This identity is derived from repeatedly applying the binomial expansion or by direct multiplication.
Explanation:
We are asked to expand (–2x + 5y – 3z)².
We can use the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
In this case, we can let:
a = –2x
b = 5y
c = –3z
Now, substitute these values into the identity:
a² = (–2x)² = 4x²
b² = (5y)² = 25y²
c² = (–3z)² = 9z²
2ab = 2(–2x)(5y) = –20xy
2bc = 2(5y)(–3z) = –30yz
2ca = 2(–3z)(–2x) = 12zx
Adding all these terms together:
(–2x + 5y – 3z)² = 4x² + 25y² + 9z² – 20xy – 30yz + 12zx
Therefore, the expanded form of (–2x + 5y – 3z)² is 4x² + 25y² + 9z² – 20xy – 30yz + 12zx.
Question:
Use suitable identity to find the following product:
(3 – 2x) (3 + 2x)
Concept in a Minute:
This question requires the application of algebraic identities. Specifically, the identity (a – b)(a + b) = a² – b² is relevant here, which represents the difference of squares. This identity is useful for simplifying the product of two binomials that have the same terms but opposite signs.
Explanation:
The given product is (3 – 2x) (3 + 2x).
We can observe that this expression is in the form of (a – b)(a + b), where a = 3 and b = 2x.
According to the difference of squares identity, (a – b)(a + b) = a² – b².
Substituting the values of a and b into the identity, we get:
(3 – 2x) (3 + 2x) = (3)² – (2x)²
Now, we calculate the squares:
(3)² = 3 * 3 = 9
(2x)² = (2x) * (2x) = 2 * 2 * x * x = 4x²
Therefore, the product is:
9 – 4x²
Final Answer: The final answer is $\boxed{9 – 4x^2}$
Question:
Following expression is polynomial in one variable or not? State reason for your answer.
x10 + y3 + t50
Concept in a Minute:
A polynomial in one variable is an algebraic expression consisting of variables and constants, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
Explanation:
The given expression is x^10 + y^3 + t^50.
A polynomial in one variable means that there should be only a single variable present in the expression, and the exponents of that variable must be non-negative integers.
In the given expression, we have three different variables: ‘x’, ‘y’, and ‘t’.
Since there are multiple variables (x, y, and t) and not just one, this expression is not a polynomial in one variable.
Therefore, the expression x^10 + y^3 + t^50 is not a polynomial in one variable. The reason is that it contains more than one variable.
Question:
Write the following cube in expanded form:
(2x + 1)3
Concept in a Minute:
The question requires expanding a cubic expression, specifically of the form (a + b)³. The relevant algebraic identity is (a + b)³ = a³ + 3a²b + 3ab² + b³.
Explanation:
To write (2x + 1)³ in expanded form, we will use the algebraic identity (a + b)³ = a³ + 3a²b + 3ab² + b³.
In this expression, ‘a’ corresponds to 2x and ‘b’ corresponds to 1.
Step 1: Identify ‘a’ and ‘b’.
Here, a = 2x and b = 1.
Step 2: Substitute ‘a’ and ‘b’ into the identity (a + b)³ = a³ + 3a²b + 3ab² + b³.
Step 3: Calculate each term:
a³ = (2x)³ = 2³ * x³ = 8x³
3a²b = 3 * (2x)² * 1 = 3 * (4x²) * 1 = 12x²
3ab² = 3 * (2x) * (1)² = 3 * (2x) * 1 = 6x
b³ = 1³ = 1
Step 4: Combine the terms to get the expanded form.
(2x + 1)³ = 8x³ + 12x² + 6x + 1
The expanded form of (2x + 1)³ is 8x³ + 12x² + 6x + 1.
Question:
Find the value of k, if x – 1 is a factor of p(x) in the following case:
p(x) = x2 + x + k
Concept in a Minute:
The Remainder Theorem states that if a polynomial p(x) is divided by a linear factor (x – a), then the remainder is p(a). A key consequence of this theorem is the Factor Theorem, which states that (x – a) is a factor of p(x) if and only if p(a) = 0.
Explanation:
The question states that (x – 1) is a factor of the polynomial p(x) = x^2 + x + k.
According to the Factor Theorem, if (x – 1) is a factor of p(x), then p(1) must be equal to 0.
So, we need to substitute x = 1 into the polynomial p(x) and set the expression equal to zero.
p(x) = x^2 + x + k
p(1) = (1)^2 + (1) + k
p(1) = 1 + 1 + k
p(1) = 2 + k
Since (x – 1) is a factor, p(1) = 0.
Therefore, 2 + k = 0.
To find the value of k, we subtract 2 from both sides of the equation:
k = -2.
Thus, the value of k is -2.
Question:
Find the zero of the polynomial in the following case:
p(x) = 2x + 5
Concept in a Minute:
A zero of a polynomial p(x) is a value of x for which p(x) = 0. To find the zero, we set the polynomial equal to zero and solve for x.
Explanation:
We are given the polynomial p(x) = 2x + 5.
To find the zero of this polynomial, we need to find the value of x for which p(x) = 0.
So, we set the polynomial equal to zero:
2x + 5 = 0
Now, we solve this linear equation for x.
First, subtract 5 from both sides of the equation:
2x + 5 – 5 = 0 – 5
2x = -5
Next, divide both sides by 2 to isolate x:
2x / 2 = -5 / 2
x = -5/2
Therefore, the zero of the polynomial p(x) = 2x + 5 is -5/2.
We can check this by substituting x = -5/2 back into the polynomial:
p(-5/2) = 2(-5/2) + 5
p(-5/2) = -5 + 5
p(-5/2) = 0
This confirms that x = -5/2 is indeed the zero of the polynomial.
Question:
Write the following cube in expanded form:
(2a – 3b)3
Concept in a Minute:
The question requires expanding a binomial cube. The relevant algebraic identity is the cube of a difference of two terms: (a – b)³ = a³ – 3a²b + 3ab² – b³.
Explanation:
We need to expand the expression (2a – 3b)³. We can use the algebraic identity (a – b)³ = a³ – 3a²b + 3ab² – b³.
In this case, a = 2a and b = 3b.
Substitute these values into the identity:
(2a – 3b)³ = (2a)³ – 3(2a)²(3b) + 3(2a)(3b)² – (3b)³
Now, let’s calculate each term:
(2a)³ = 2³ * a³ = 8a³
3(2a)²(3b) = 3(4a²)(3b) = 3 * 4 * 3 * a² * b = 36a²b
3(2a)(3b)² = 3(2a)(9b²) = 3 * 2 * 9 * a * b² = 54ab²
(3b)³ = 3³ * b³ = 27b³
Now, combine these terms with the appropriate signs:
(2a – 3b)³ = 8a³ – 36a²b + 54ab² – 27b³
Therefore, the expanded form of (2a – 3b)³ is 8a³ – 36a²b + 54ab² – 27b³.
Question:
Determine the following polynomial has (x + 1) a factor:
x4 + x3 + x2 + x + 1
Concept in a Minute:
The Remainder Theorem states that if a polynomial P(x) is divided by (x – a), then the remainder is P(a). If (x + 1) is a factor of a polynomial P(x), then the remainder when P(x) is divided by (x + 1) must be zero. This means that P(-1) must be equal to 0.
Explanation:
We are given the polynomial P(x) = x⁴ + x³ + x² + x + 1.
To determine if (x + 1) is a factor, we need to check if P(-1) = 0.
Substitute x = -1 into the polynomial:
P(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
P(-1) = 1 + (-1) + 1 + (-1) + 1
P(-1) = 1 – 1 + 1 – 1 + 1
P(-1) = 0 + 1 – 1 + 1
P(-1) = 1 + 0 + 1
P(-1) = 2
Since P(-1) = 2, which is not equal to 0, (x + 1) is not a factor of the given polynomial.
Therefore, the polynomial x⁴ + x³ + x² + x + 1 does not have (x + 1) as a factor.
Question:
Classify the following as linear, quadratic and cubic polynomial:
3t
Concept in a Minute:
A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.
The degree of a polynomial is the highest exponent of the variable in the polynomial.
A linear polynomial has a degree of 1.
A quadratic polynomial has a degree of 2.
A cubic polynomial has a degree of 3.
Explanation:
The given polynomial is 3t.
In this polynomial, the variable is ‘t’.
The exponent of the variable ‘t’ is 1 (since t can be written as t^1).
Since the highest exponent of the variable is 1, the degree of the polynomial is 1.
A polynomial with a degree of 1 is classified as a linear polynomial.
Therefore, 3t is a linear polynomial.
Question:
Factorise the following:
27y3 + 125z3
Concept in a Minute:
This question requires the application of the sum of cubes factorization formula: a³ + b³ = (a + b)(a² – ab + b²).
Explanation:
The given expression is 27y³ + 125z³.
We need to identify ‘a’ and ‘b’ such that a³ = 27y³ and b³ = 125z³.
We know that 3³ = 27, so a³ = (3y)³. Therefore, a = 3y.
We know that 5³ = 125, so b³ = (5z)³. Therefore, b = 5z.
Now, we apply the sum of cubes formula: a³ + b³ = (a + b)(a² – ab + b²).
Substitute a = 3y and b = 5z into the formula:
(3y)³ + (5z)³ = (3y + 5z)((3y)² – (3y)(5z) + (5z)²)
Simplify the terms within the second bracket:
(3y)² = 9y²
(3y)(5z) = 15yz
(5z)² = 25z²
So, the factorization becomes:
(3y + 5z)(9y² – 15yz + 25z²)
Therefore, the factorized form of 27y³ + 125z³ is (3y + 5z)(9y² – 15yz + 25z²).
Question:
Verify whether the following zeroes of the polynomial, indicated against them.
p(x) = (x + 1) (x – 2), x = – 1, 2
Concept in a Minute:
A zero of a polynomial p(x) is a value of x for which p(x) = 0. To verify if a given value is a zero, substitute that value into the polynomial and check if the result is zero.
Explanation:
We are given the polynomial p(x) = (x + 1)(x – 2) and are asked to verify if x = – 1 and x = 2 are its zeroes.
To verify if x = – 1 is a zero, we substitute x = – 1 into the polynomial:
p(– 1) = (– 1 + 1)(– 1 – 2)
p(– 1) = (0)(– 3)
p(– 1) = 0
Since p(– 1) = 0, x = – 1 is a zero of the polynomial.
To verify if x = 2 is a zero, we substitute x = 2 into the polynomial:
p(2) = (2 + 1)(2 – 2)
p(2) = (3)(0)
p(2) = 0
Since p(2) = 0, x = 2 is a zero of the polynomial.
Therefore, both x = – 1 and x = 2 are zeroes of the polynomial p(x) = (x + 1)(x – 2).
Question:
Find p(0), p(1) and p(2) for the following polynomial:-
p(x) = x3
Concept in a Minute:
A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. To find the value of a polynomial at a specific point, substitute the given value for the variable and evaluate the expression.
Explanation:
The given polynomial is p(x) = x³. We need to find the values of p(0), p(1), and p(2).
To find p(0), substitute x = 0 into the polynomial:
p(0) = (0)³
p(0) = 0 * 0 * 0
p(0) = 0
To find p(1), substitute x = 1 into the polynomial:
p(1) = (1)³
p(1) = 1 * 1 * 1
p(1) = 1
To find p(2), substitute x = 2 into the polynomial:
p(2) = (2)³
p(2) = 2 * 2 * 2
p(2) = 8
Therefore, p(0) = 0, p(1) = 1, and p(2) = 8.
Question:
Find p(0), p(1) and p(2) for the following polynomial:-
p(t) = 2 + t + 2t2 – t3
Concept in a Minute:
Evaluating a polynomial means substituting a given value for the variable and simplifying the expression.
Explanation:
To find p(0), p(1), and p(2), we will substitute these values for ‘t’ in the given polynomial p(t) = 2 + t + 2t² – t³.
For p(0):
Substitute t = 0 into the polynomial:
p(0) = 2 + (0) + 2(0)² – (0)³
p(0) = 2 + 0 + 2(0) – 0
p(0) = 2 + 0 + 0 – 0
p(0) = 2
For p(1):
Substitute t = 1 into the polynomial:
p(1) = 2 + (1) + 2(1)² – (1)³
p(1) = 2 + 1 + 2(1) – 1
p(1) = 2 + 1 + 2 – 1
p(1) = 3 + 2 – 1
p(1) = 5 – 1
p(1) = 4
For p(2):
Substitute t = 2 into the polynomial:
p(2) = 2 + (2) + 2(2)² – (2)³
p(2) = 2 + 2 + 2(4) – 8
p(2) = 2 + 2 + 8 – 8
p(2) = 4 + 8 – 8
p(2) = 12 – 8
p(2) = 4
Therefore:
p(0) = 2
p(1) = 4
p(2) = 4
Question:
Expand the following, using suitable identity:
(3a – 7b – c)2
Concept in a Minute:
The question requires the expansion of a trinomial squared. The suitable identity to use is the square of a trinomial: (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx. It’s important to correctly identify ‘x’, ‘y’, and ‘z’ from the given expression, paying attention to their signs.
Explanation:
The given expression is (3a – 7b – c)². We can use the identity (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.
In this case, let x = 3a, y = -7b, and z = -c.
Now, substitute these values into the identity:
x² = (3a)² = 9a²
y² = (-7b)² = 49b²
z² = (-c)² = c²
2xy = 2(3a)(-7b) = -42ab
2yz = 2(-7b)(-c) = 14bc
2zx = 2(-c)(3a) = -6ca
Adding all these terms together:
(3a – 7b – c)² = 9a² + 49b² + c² – 42ab + 14bc – 6ca
Question:
Classify the following as linear, quadratic and cubic polynomial:
1 + x
Concept in a Minute:
A polynomial is classified by the highest power (degree) of the variable present in the expression.
– Linear polynomial: The highest power of the variable is 1. (e.g., ax + b)
– Quadratic polynomial: The highest power of the variable is 2. (e.g., ax^2 + bx + c)
– Cubic polynomial: The highest power of the variable is 3. (e.g., ax^3 + bx^2 + cx + d)
Explanation:
The given polynomial is 1 + x.
We can rewrite this as x + 1.
The variable in this polynomial is ‘x’.
The highest power of ‘x’ in the expression is 1 (since x is the same as x^1).
Therefore, according to the classification of polynomials, a polynomial with the highest power of the variable as 1 is a linear polynomial.
Answer:
Linear polynomial
Question:
Without actually calculating the cubes, find the value of the following:
(28)3 + (–15)3 + (–13)3
Concept in a Minute:
Identity of sum of cubes: If a + b + c = 0, then a³ + b³ + c³ = 3abc.
Explanation:
Let a = 28, b = -15, and c = -13.
First, check if the sum of these numbers is zero:
a + b + c = 28 + (–15) + (–13)
a + b + c = 28 – 15 – 13
a + b + c = 28 – 28
a + b + c = 0
Since a + b + c = 0, we can use the identity a³ + b³ + c³ = 3abc.
Substitute the values of a, b, and c into the identity:
(28)³ + (–15)³ + (–13)³ = 3 * (28) * (–15) * (–13)
Now, perform the multiplication:
First, multiply 3 by 28:
3 * 28 = 84
Next, multiply –15 by –13. Since both numbers are negative, the result will be positive:
–15 * –13 = 195
Finally, multiply 84 by 195:
84 * 195 = 16380
Therefore, (28)³ + (–15)³ + (–13)³ = 16380.
Question:
Evaluate the following product without multiplying directly:
103 × 107
Concept in a Minute:
The question requires evaluating a product without direct multiplication. This suggests using algebraic identities to simplify the calculation. Specifically, the identity (a + b)(a + c) = a^2 + (b + c)a + bc is useful when dealing with numbers close to a round number like 100.
Explanation:
We need to evaluate 103 × 107 without direct multiplication.
We can rewrite the numbers as (100 + 3) and (100 + 7).
This fits the pattern of the algebraic identity (a + b)(a + c) = a^2 + (b + c)a + bc.
In this case, let a = 100, b = 3, and c = 7.
Applying the identity:
(100 + 3)(100 + 7) = 100^2 + (3 + 7) × 100 + (3 × 7)
= 10000 + (10) × 100 + 21
= 10000 + 1000 + 21
= 11021
Therefore, 103 × 107 = 11021.
Question:
Determine the following polynomial has (x + 1) a factor:
x3 + x2 + x + 1
Concept in a Minute:
The Factor Theorem states that a polynomial P(x) has a factor (x – a) if and only if P(a) = 0. In this case, we are checking if (x + 1) is a factor, which is equivalent to checking if (x – (-1)) is a factor. Therefore, we need to evaluate the polynomial at x = -1.
Explanation:
Let the given polynomial be P(x) = x³ + x² + x + 1.
To determine if (x + 1) is a factor of P(x), we apply the Factor Theorem.
According to the Factor Theorem, (x + 1) is a factor of P(x) if P(-1) = 0.
Substitute x = -1 into the polynomial P(x):
P(-1) = (-1)³ + (-1)² + (-1) + 1
P(-1) = -1 + 1 – 1 + 1
P(-1) = 0
Since P(-1) = 0, by the Factor Theorem, (x + 1) is a factor of the polynomial x³ + x² + x + 1.
Question:
Classify the following as linear, quadratic, and cubic polynomial:
x – x3
Concept in a Minute:
The degree of a polynomial is the highest power of the variable in the polynomial.
A linear polynomial has a degree of 1.
A quadratic polynomial has a degree of 2.
A cubic polynomial has a degree of 3.
Explanation:
The given polynomial is x – x³.
To classify this polynomial, we need to determine its degree.
The terms in the polynomial are ‘x’ and ‘-x³’.
The power of ‘x’ in the first term is 1 (since x = x¹).
The power of ‘x’ in the second term is 3.
The highest power of the variable ‘x’ in the polynomial is 3.
Therefore, the degree of the polynomial x – x³ is 3.
A polynomial with a degree of 3 is classified as a cubic polynomial.
Final Classification: Cubic polynomial
Question:
Find the value of k, if x – 1 is a factor of p(x) in the following case:
p(x) = kx2 – 3x + k
Concept in a Minute:
The Factor Theorem states that if (x – a) is a factor of a polynomial p(x), then p(a) = 0.
Explanation:
We are given that (x – 1) is a factor of the polynomial p(x) = kx² – 3x + k.
According to the Factor Theorem, if (x – 1) is a factor of p(x), then p(1) must be equal to 0.
To find p(1), we substitute x = 1 into the polynomial p(x):
p(1) = k(1)² – 3(1) + k
Now, we simplify the expression:
p(1) = k(1) – 3 + k
p(1) = k – 3 + k
Combine the like terms:
p(1) = 2k – 3
Since p(1) = 0, we can set up the equation:
2k – 3 = 0
Now, we solve for k:
Add 3 to both sides of the equation:
2k = 3
Divide both sides by 2:
k = 3/2
Therefore, the value of k is 3/2.
Question:
Find p(0), p(1) and p(2) for the following polynomial:-
p(x) = (x – 1) (x + 1)
Concept in a Minute:
To find the value of a polynomial p(x) for a specific value of x, substitute that value into the polynomial expression and simplify. This is also known as evaluating the polynomial at a point.
Explanation:
The given polynomial is p(x) = (x – 1) (x + 1).
We need to find p(0), p(1), and p(2).
To find p(0), substitute x = 0 into the polynomial:
p(0) = (0 – 1) (0 + 1)
p(0) = (-1) (1)
p(0) = -1
To find p(1), substitute x = 1 into the polynomial:
p(1) = (1 – 1) (1 + 1)
p(1) = (0) (2)
p(1) = 0
To find p(2), substitute x = 2 into the polynomial:
p(2) = (2 – 1) (2 + 1)
p(2) = (1) (3)
p(2) = 3
Therefore, p(0) = -1, p(1) = 0, and p(2) = 3.
Question:
Classify the following as linear, quadratic and cubic polynomial:
y + y2 + 4
Concept in a Minute:
A polynomial is classified based on the highest power (degree) of its variable.
Linear polynomial: Degree is 1.
Quadratic polynomial: Degree is 2.
Cubic polynomial: Degree is 3.
Explanation:
The given polynomial is y + y^2 + 4.
To classify this polynomial, we need to identify the highest power of the variable ‘y’.
The terms in the polynomial are ‘y’ (which has a power of 1), ‘y^2’ (which has a power of 2), and ‘4’ (which is a constant term and can be thought of as y^0).
The powers of ‘y’ are 1, 2, and 0.
The highest power among these is 2.
Since the highest power of the variable is 2, the polynomial is a quadratic polynomial.
Question:
Find the zero of the polynomial in the following case:
p(x) = cx + d, c ≠ 0, c, d are real numbers.
Concept in a Minute:
The zero of a polynomial is the value of the variable for which the polynomial evaluates to zero. For a linear polynomial of the form p(x) = ax + b, setting p(x) = 0 and solving for x gives the zero.
Explanation:
To find the zero of the polynomial p(x) = cx + d, we need to find the value of x for which p(x) = 0.
So, we set the polynomial equal to zero:
cx + d = 0
Now, we need to solve this equation for x.
First, subtract d from both sides of the equation:
cx = -d
Since we are given that c ≠ 0, we can divide both sides of the equation by c to isolate x:
x = -d/c
Therefore, the zero of the polynomial p(x) = cx + d is -d/c.
Question:
Find the zero of the polynomial in the following case:
p(x) = x – 5
Concept in a Minute:
To find the zero of a polynomial, we set the polynomial equal to zero and solve for the variable. A zero of a polynomial p(x) is a value of x for which p(x) = 0.
Explanation:
The given polynomial is p(x) = x – 5.
To find the zero of this polynomial, we need to find the value of x for which p(x) = 0.
So, we set the polynomial equal to zero:
x – 5 = 0
To solve for x, we add 5 to both sides of the equation:
x – 5 + 5 = 0 + 5
x = 5
Therefore, the zero of the polynomial p(x) = x – 5 is 5.
We can verify this by substituting x = 5 back into the polynomial:
p(5) = 5 – 5 = 0.
Since p(5) = 0, our answer is correct.
Question:
Without actually calculating the cubes, find the value of the following:
(–12)3 + (7)3 + (5)3
Concept in a Minute:
The problem requires the use of the algebraic identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc. This identity helps to find the sum of cubes without direct calculation, provided the sum of the bases is zero.
Explanation:
Let a = -12, b = 7, and c = 5.
First, we check the sum of the bases:
a + b + c = (-12) + 7 + 5
a + b + c = -12 + 12
a + b + c = 0
Since the sum of the bases is 0, we can apply the identity a³ + b³ + c³ = 3abc.
Therefore, (-12)³ + (7)³ + (5)³ = 3 * (-12) * 7 * 5.
Now, we calculate the product:
3 * (-12) = -36
-36 * 7 = -252
-252 * 5 = -1260
So, (-12)³ + (7)³ + (5)³ = -1260.
Question:
Classify the following as linear, quadratic and cubic polynomial:
x2 + x
Concept in a Minute:
The degree of a polynomial is the highest power of the variable in the polynomial.
A polynomial with degree 1 is called a linear polynomial.
A polynomial with degree 2 is called a quadratic polynomial.
A polynomial with degree 3 is called a cubic polynomial.
Explanation:
The given polynomial is x^2 + x.
In this polynomial, the terms are x^2 and x.
The powers of the variable x in these terms are 2 and 1, respectively.
The highest power of x in the polynomial is 2.
Therefore, the degree of the polynomial x^2 + x is 2.
A polynomial with a degree of 2 is classified as a quadratic polynomial.
Question:
Find p(0), p(1) and p(2) for the following polynomial:-
p(y) = y2 – y + 1
Concept in a Minute:
Evaluating a polynomial means substituting a given value for the variable in the polynomial expression and calculating the resulting value.
Explanation:
The given polynomial is p(y) = y^2 – y + 1.
To find p(0), we substitute y = 0 into the polynomial:
p(0) = (0)^2 – (0) + 1
p(0) = 0 – 0 + 1
p(0) = 1
To find p(1), we substitute y = 1 into the polynomial:
p(1) = (1)^2 – (1) + 1
p(1) = 1 – 1 + 1
p(1) = 1
To find p(2), we substitute y = 2 into the polynomial:
p(2) = (2)^2 – (2) + 1
p(2) = 4 – 2 + 1
p(2) = 2 + 1
p(2) = 3
Therefore, p(0) = 1, p(1) = 1, and p(2) = 3.
Question:
Write the degree of the following polynomial:-
4 – y2
Concept in a Minute:
The degree of a polynomial is the highest power of the variable present in the polynomial.
Explanation:
The given polynomial is 4 – y^2.
This polynomial has one variable, which is ‘y’.
We need to identify the power of ‘y’ in each term.
The first term is ‘4’. We can consider this as 4 * y^0, so the power of ‘y’ is 0.
The second term is ‘-y^2’. The power of ‘y’ in this term is 2.
The degree of the polynomial is the highest power of the variable. Comparing the powers 0 and 2, the highest power is 2.
Therefore, the degree of the polynomial 4 – y^2 is 2.
Degree: 2
Question:
Find the zero of the polynomial in the following case:
p(x) = x + 5
Concept in a Minute:
To find the zero of a polynomial, we need to find the value of the variable (in this case, x) for which the polynomial evaluates to zero. This means setting the polynomial equal to zero and solving for the variable.
Explanation:
The given polynomial is p(x) = x + 5.
To find the zero of this polynomial, we need to find the value of x for which p(x) = 0.
So, we set the polynomial equal to zero:
x + 5 = 0
To solve for x, we subtract 5 from both sides of the equation:
x + 5 – 5 = 0 – 5
x = -5
Therefore, the zero of the polynomial p(x) = x + 5 is -5.
Question:
Classify the following as linear, quadratic and cubic polynomial:
r2
Concept in a Minute:
The degree of a polynomial is the highest exponent of the variable in the polynomial. A polynomial is classified based on its degree:
– Linear polynomial: Degree 1
– Quadratic polynomial: Degree 2
– Cubic polynomial: Degree 3
Explanation:
The given polynomial is r^2.
The variable in this polynomial is ‘r’.
The exponent of the variable ‘r’ is 2.
Since the highest exponent of the variable is 2, the degree of the polynomial is 2.
A polynomial with a degree of 2 is classified as a quadratic polynomial.
Therefore, r^2 is a quadratic polynomial.
Question:
Find the zero of the polynomial in the following case:
p(x) = 3x – 2
Concept in a Minute:
A zero of a polynomial p(x) is the value of x for which p(x) = 0. To find the zero, we set the polynomial equal to zero and solve for x.
Explanation:
We are given the polynomial p(x) = 3x – 2.
To find the zero of this polynomial, we need to find the value of x for which p(x) = 0.
So, we set the polynomial equal to zero:
3x – 2 = 0
Now, we solve this linear equation for x.
Add 2 to both sides of the equation:
3x – 2 + 2 = 0 + 2
3x = 2
Divide both sides by 3:
3x / 3 = 2 / 3
x = 2/3
Therefore, the zero of the polynomial p(x) = 3x – 2 is 2/3.
Question:
Write the degree of the following polynomial:-
3
Concept in a Minute:
The degree of a polynomial is the highest power of the variable present in the polynomial. A constant term, like ‘3’ in this case, can be considered a polynomial where the variable’s power is 0, as any non-zero number raised to the power of 0 is 1.
Explanation:
The given expression is ‘3’. This is a constant polynomial. A constant polynomial can be written as $3x^0$, because $x^0 = 1$ for any non-zero $x$. The power of the variable $x$ in this expression is 0. Therefore, the degree of the polynomial ‘3’ is 0.
Question:
Write the degree of the following polynomial:-
5x3 + 4x2 + 7x
Concept in a Minute:
The degree of a polynomial is the highest power of the variable present in the polynomial.
Explanation:
The given polynomial is 5x³ + 4x² + 7x.
We need to identify the powers of the variable ‘x’ in each term.
In the first term, 5x³, the power of x is 3.
In the second term, 4x², the power of x is 2.
In the third term, 7x, the power of x is 1 (since x can be written as x¹).
The highest power among these is 3.
Therefore, the degree of the polynomial 5x³ + 4x² + 7x is 3.
Degree: 3
Question:
Verify whether the following zeroes of the polynomial, indicated against them.
p(x) = x2, x = 0
Concept in a Minute:
To verify if a given value is a zero of a polynomial, substitute the value into the polynomial. If the result is zero, then the value is indeed a zero of the polynomial.
Explanation:
The given polynomial is p(x) = x^2.
We need to verify if x = 0 is a zero of this polynomial.
To do this, we substitute x = 0 into the polynomial p(x).
p(0) = (0)^2
p(0) = 0
Since p(0) = 0, it means that x = 0 is a zero of the polynomial p(x) = x^2.
Question:
Find the value of the polynomial 5x – 4x2 + 3 at x = 0.
Concept in a Minute:
Evaluating a polynomial involves substituting a given value for the variable(s) and performing the arithmetic operations.
Explanation:
The given polynomial is P(x) = 5x – 4x² + 3.
We are asked to find the value of this polynomial at x = 0.
To do this, we substitute x = 0 into the polynomial:
P(0) = 5(0) – 4(0)² + 3
P(0) = 0 – 4(0) + 3
P(0) = 0 – 0 + 3
P(0) = 3
Therefore, the value of the polynomial 5x – 4x² + 3 at x = 0 is 3.
Question:
Find the zero of the polynomial in the following case:
p(x) = 3x
Concept in a Minute:
A zero of a polynomial p(x) is a value of x for which p(x) = 0. To find the zero, we set the polynomial equal to zero and solve for x.
Explanation:
To find the zero of the polynomial p(x) = 3x, we need to find the value of x for which p(x) equals 0.
So, we set the polynomial equal to zero:
3x = 0
To solve for x, we divide both sides of the equation by 3:
x = 0 / 3
x = 0
Therefore, the zero of the polynomial p(x) = 3x is 0.
Question:
Classify the following as linear, quadratic and cubic polynomial:
7x3
Concept in a Minute:
A polynomial is classified based on the highest power (degree) of the variable.
Linear polynomial: Degree is 1.
Quadratic polynomial: Degree is 2.
Cubic polynomial: Degree is 3.
Explanation:
The given polynomial is 7x^3.
The variable in this polynomial is ‘x’.
The highest power of ‘x’ is 3.
Therefore, the degree of the polynomial is 3.
A polynomial with a degree of 3 is classified as a cubic polynomial.
Next Chapter: Quadrilaterals
Refer Polynomials Notes
Practice Polynomials Extra Questions
Conquer Maths & Science – with LearnTheta’s AI-Practice!
✅ All Topics at One Place
🤖 Adaptive Question Practice
📊 Progress and Insights